# How do you differentiate #y=tanx/(2x^3)#?

##### 1 Answer

Dec 12, 2016

#### Explanation:

If you are studying maths, then you should learn the Quotient Rule for Differentiation, and practice how to use it:

# d/dx(u/v) = (v(du)/dx-u(dv)/dx)/v^2 # , or less formally,# (u/v)' = (v(du)-u(dv))/v^2 #

I was taught to remember the rule in word; " *vdu minus udv all over v squared* ". To help with the ordering I was taught to remember the acronym, VDU as in Visual Display Unit.

So with

# :. d/dx(u/v) = (v(du)/dx-u(dv)/dx)/v^2 #

# :. " " dy/dx = ( (2x^3)(sec^2x) - (tanx)(6x^2) ) / (2x^3)^2#

# :. " " dy/dx = ( 2x^3sec^2x - 6x^2tanx ) / (4x^6)#

# :. " " dy/dx = ( x^2sec^2x - 3tanx ) / (2x^4)#